Answer:
Step-by-step explanation:
This is a differential equation problem most easily solved with an exponential decay equation of the form
[tex]y=Ce^{kt}[/tex]. We know that the initial amount of salt in the tank is 28 pounds, so
C = 28. Now we just need to find k.
The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is [tex]\frac{dy}{dt}[/tex]. Thus, the change in the concentration of salt is found in
[tex]\frac{dy}{dt}=[/tex] inflow of salt - outflow of salt
Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:
[tex]3(\frac{y}{400})[/tex]
Therefore,
[tex]\frac{dy}{dt}=0-3(\frac{y}{400})[/tex] or just
[tex]\frac{dy}{dt}=-\frac{3y}{400}[/tex] and in terms of time,
[tex]-\frac{3t}{400}[/tex]
Thus, our equation is
[tex]y=28e^{-\frac{3t}{400}[/tex] and filling in 16 for the number of minutes in t:
y = 24.834 pounds of salt
